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2 years ago

Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.

Physics

1 answer:

AleksandrR [38]2 years ago

5 0

Answer:

$q=390.285x10^{-12}C$

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

$7.0cm*\frac{1m}{100cm}=70x10^{-3}m$

$A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62$

Capacitance

$C=E_o*\frac{A}{d}$

$E_o=8.85x10^{-12}\frac{C^2}{N*m^2}$

$C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}$

$C=43.365x10^{-12}F$

$C=43.365pF$

The charge is find by the equation

$q=C*V$

$q=43.365pF*9V$

$q=390.285x10^{-12}C$

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What volume in milliliters will 0.00922 g of h2 gas occupy at stp?

Nadusha1986 [10]

Assuming that this gas is in ideal state, we can use the relation that for every 1 mol of an ideal gas it would have a volume of 22.4 L. But before using this, relation we need to convert the number of grams of H2 into moles by using the molar mass of 2.02 g/mol.

moles H2 = 0.00922 g ( 1 mol / 2.02 g ) = 0.005 mol H2

Volume H2 at STP = 0.005 mol H2 ( 22.4 L / 1 mol ) = 0.102 L of H2

4 0

2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

2 years ago

A car wheel turns through 277° in 10.7 s. Calculate the angular speed of the wheel.

slava [35]

Answer:

The angular speed of the wheel is 0.452 rad/s

Explanation:

The angle through which the car wheel turns, Δθ = 277° = 277/360 × 2·π radian

The time it takes for the car wheel to turn, Δt = 10.7 s

The angular speed, ω is given by the following equation;

$Angular \ speed = \dfrac{Change \ in \ angular \ rotation }{Change \ in \ time} = \dfrac{\Delta \theta}{\Delta t}$

Substituting the known values for Δθ and Δt gives;

$Angular \ speed = \dfrac{\dfrac{277 ^{\circ}}{360 ^{\circ } } \times 2 \times \pi \ radian}{10.7 \ seconds} \approx 0.452 \ rad/s$

The angular speed of the wheel = 0.452 rad/s

3 0

2 years ago

Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿

Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0

2 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$